Question

[Dángksmsmms]

A=1/2+1/2^2+1/2^3+1/2^4+...+1/2^100.

Chứng tỏ: A<1

Answer 1

[giang ho dai ca]

2A= \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)

=> 2A - A= \(\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{^{2^{100}}}\right)\)

=> A = \(1-\frac{1}{2^{100}}\)< 1

=> A< 1

đúng nhé

Answer 2

[Katherine Lilly Filbert]

A=\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{100}}\)

2A=\(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}\)

2A-A=\(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}\)\(-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{100}}\right)\)

A=\(1-\frac{1}{2^{100}}\)

Vì \(1-\frac{1}{2^{100}}\)< \(1\)

Nên \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{100}}\)< \(1\)

Vậy A <\(1\)